Indefinite Integration 1 Question 35
36. If $f(x)=A \sin \frac{\pi x}{2}+B$,
$f^{\prime} \frac{1}{2}=\sqrt{2}$ and $\int _0^{1} f(x) d x=\frac{2 A}{\pi}$, then constants
$A$ and $B$ are
(1995, 2M)
(a) $\frac{\pi}{2}$ and $\frac{\pi}{2}$
(b) $\frac{2}{\pi}$ and $\frac{3}{\pi}$
(c) 0 and $-\frac{4}{\pi}$
(d) $\frac{4}{\pi}$ and 0
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Solution:
- Given, $f(x)=A \sin \frac{\pi x}{2}+B, f^{\prime} \frac{1}{2}=\sqrt{2}$
$$ \text { and } \quad \int _0^{1} f(x) d x=\frac{2 A}{\pi} $$
$$ f^{\prime}(x)=\frac{A \pi}{2} \cos \frac{\pi x}{2} \Rightarrow f^{\prime} \frac{1}{2}=\frac{A \pi}{2} \cos \frac{\pi}{4}=\frac{A \pi}{2 \sqrt{2}} $$
$$ \text { But } \quad f^{\prime} \frac{1}{2}=\sqrt{2} \therefore \quad \frac{A \pi}{2 \sqrt{2}}=\sqrt{2} \quad \Rightarrow \quad A=\frac{4}{\pi} $$
Now, $\int _0^{1} f(x) d x=\frac{2 A}{\pi} \Rightarrow \int _0^{1} A \sin \frac{\pi x}{2}+B d x=\frac{2 A}{\pi}$
$$ \begin{aligned} \Rightarrow & -\frac{2 A}{\pi} \cos \frac{\pi x}{2}+B x & =\frac{2 A}{\pi} \Rightarrow B+\frac{2 A}{\pi}=\frac{2 A}{\pi} \\ \Rightarrow & B & =0 \end{aligned} $$
